# transformation of singular quartics into Weierstrass form

I need a birational transformation of singular quartics into weierstrass form. the quartics are of genus one and have the following form: $$(x^2 + c) (y^2 + i) + k = 0$$ where $$c,i,k$$ are given rational numbers, and $$k$$ is of big height. The only known rational solutions: the singular points in homogenous coordinates: $$(x,y,z)=(0,1,0),(1,0,0)$$.

Methods described in the literature (e.g. Lawrence Washington, Elliptic curves, p.37) require a non-singular rational solution. but after this text a singular quartic can be transformed birationally into a nonsingular curve, where the singular solutions are mapped into two nonsingular solutions. Unfortunately, I have not found in this work and in google, how to do this transformation. I ask for information, how to transform singular quartics as above into Weierstrass form or how to transform into a nonsingular curve.

• Welcome to MSE. You'll get a lot better response if you take the time to format your post using MathJax. For some basic information about writing mathematics at this site see, basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. Jul 31 '20 at 18:24
• All the algorithms I know of for this problem require you to have a smooth point I think that it is nontrivial to find a rational one in general - of course it is easy to find one over the algebraic closure. If you are given your $c, i, k$ then Magma's "PointSearch" should look for you. Another trick to try would be to compute the blow up at the singular point and pray that there is an obvious point there - i.e., if you have rational slopes at the singular point. Sadly this doesn't seem to be the case for you. Aug 1 '20 at 0:30
• Do you want to do this transformation over $\mathbb{Q}$? If so, I'm not sure that it's always possible. There are genus $1$ curves over $\mathbb{Q}$ with no rational points, and which therefore cannot be put into Weierstrass form over $\mathbb{Q}$. If you don't mind extending the base field, then finding a smooth point is easy: just set $x=0$ (or some other constant) and solve for $y$, or vice versa. Aug 1 '20 at 3:50

This is just an extened comment. I would conjecture that it is unlikely that there is a nice closed from for a Weierstrass equation for the curve $$C: (X^2 + rZ^2)(Y^2 + sZ^2) + tZ^4 = 0$$

over the function field $$\mathbb{Q}(r,s,t)$$ - maybe some expert on elliptic surfaces will come along and say that this is codswhallop.

Why do I say this? Let's ask Magma about the case where $$r = -u^2$$ (we will see why this is helpful). Take the affine piece where $$y = 1$$ and compute the blowup (as in e.g., Hartshorne I Example 4.9.1) see that the curve $$C' : sx^2z^2 + x^2 + (rs + t)z^2 + r = 0$$ is birational to $$C$$. Notice that $$C'$$ has the nonsingular point $$(u, 0)$$.

Now let's ask Magma what the Weierstrass equation is, shall we?

K<r,s,t> := FunctionField(Rationals(), 3);

A2<x,z> := AffineSpace(K, 2);

f := (x^2 -r^2*z^2)*(1 + s*z^2) + t*z^4;
C:= Curve(A2, f);

g := ExactQuotient(Evaluate(f, [x*z, z]), z^2);

C1 := Curve(A2, g);
C1_proj := ProjectiveClosure(C1);
P := C1_proj![r, 0, 1];
IsSingular(P);                            //Should be false for the next bit to work

E := EllipticCurve(C1_proj, P);
WeierstrassModel(E);


Which outputs the lovely equation

Elliptic Curve defined by y^2 = x^3 + (-6912*r^8*s^6 + 6912*r^6*s^5*t -
432*r^4*s^4*t^2)/(r^16*s^8 - 6*r^14*s^7*t + 31/2*r^12*s^6*t^2 -
45/2*r^10*s^5*t^3 + 321/16*r^8*s^4*t^4 - 45/4*r^6*s^3*t^5 + 31/8*r^4*s^2*t^6
- 3/4*r^2*s*t^7 + 1/16*t^8)*x + (-221184*r^10*s^8 + 221184*r^8*s^7*t +
6912*r^6*s^6*t^2)/(r^22*s^11 - 17/2*r^20*s^10*t + 65/2*r^18*s^9*t^2 -
295/4*r^16*s^8*t^3 + 1765/16*r^14*s^7*t^4 - 3653/32*r^12*s^6*t^5 +
667/8*r^10*s^5*t^6 - 1375/32*r^8*s^4*t^7 + 245/16*r^6*s^3*t^8 -
115/32*r^4*s^2*t^9 + 1/2*r^2*s*t^10 - 1/32*t^11) over Multivariate rational
function field of rank 3 over Rational Field